4 Answers
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The standard example is an interval and a circle. Consider the interval $[0,2\pi)$ and the circle $S^1 = \{(x,y) \in \mathbb{R}^2 \, | \, x^2 + y^2 = 1 \}$ (with the induced topology as a subset of $\mathbb{R}^2$). You have a continuous map $\varphi \colon [0,2\pi) \rightarrow S^1$ given by $\varphi(\theta) = (\cos(\theta), \sin(\theta))$. This map is one to one and onto, but it doesn't have a continuous inverse.

The inverse map $\varphi^{-1} \colon S^1 \rightarrow [0,2\pi)$ takes a point $(x,y)$ on the circle and returns the angle the vector $(x,y)$ makes with the $x$-axis, in radians, in the range $[0,2\pi)$. This map is not continuous because if you look approach the point $(0,1)$ on the circle using points that have a positive $y$-coordinate, the angle $\varphi^{-1}$ will approach $0$ while if you approach the point $(0,1)$ with points having a negative $y$-coordinate, the angle $\varphi^{-1}$ will "approach $2\pi$" so the "left limit" and the "right limit" at $(0,1)$ are not equal for $\varphi^{-1}$.

(This is not entirely rigorous because we consider the map $\varphi^{-1}$ as a map with codomain $[0,2\pi)$ so $2\pi$ is not in the codomain and in fact when we approach $(0,1)$ with points having a negative $y$-coordinate, the angle $\varphi^{-1}$ won't have a limit in $[0,2\pi)$ at all! However, this is enough to show that $\varphi^{-1}$ can't be continuous).

$\begingroup$Can we also use this trick to prove that there is no homeomorphism between a circle and $\mathbb{R}$? But a circle is still a 1-dimensional manifold because we can choose a different homeomorphism at various points of the circle?$\endgroup$
– Alexandre Holden DalyJul 17 '14 at 20:33

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$\begingroup$I only argued that $\varphi$ is not a homeomorphism, not that other homeomorphisms aren't possible. This is also true - a topological space $X$ is called sequentially compact if every sequence $(x_n)$ of points in $X$ has a convergent subsequence. This is a topological notion and so preserved by homeomorphisms - if a space is sequentially compact then any space homeomorphic to it is also sequentially compact. The circle $S^1$ is sequentially compact while both $[0,1)$ and $\mathbb{R}$ aren't. Finally, like you wrote, locally, $\mathbb{R}$ and $S^1$ are homeomorphic.$\endgroup$
– levapJul 17 '14 at 21:29

A generic way to obtain examples of bijective continuous functions whose inverse is not continuous is to fix some set $X$ and put two different topologies $T_1$ and $T_2$ on $X$, such that $T_2\subset T_1$. Now the identity function from $X$ to itself (with the topology being $T_1$ on the domain and $T_2$ on the codomain) will be continuous and bijective, but unless $T_1 = T_2$ it will not be a homeomorphism.

Constant functions are continuous. So consider some space that has more than one point, such the circle $S = \{(\cos \theta, \sin\theta) \mid \theta\in\Bbb R\}$, considered as a subspace of $\Bbb R^2$. A mapping $f:S\to\{x\}$ is continuous, but is not a homeomorphism, because it is not a bijection. And indeed the circle is not homeomorphic to a single point.

Now consider the mapping from the circle $S$ to the closed interval $[-1,1]$ where each point $(x, y)$ is mapped onto the value $y$. This mapping is continuous but is not a homeomorphism (it is not invertible) and indeed the circle is not homeomorphic to the closed interval $[-1,1]$.

For an example which is a bijection, consider the mapping $f$ from the space $X = [0, 1) \cup [2,3]$, considered as a subset of the real line, defined as follows: $$f(x) = \begin{cases} x & \text{if $x< 1$} \\
x-1 & \text{if $x>1$}\end{cases}$$

This is a continuous bijection from $X$ to the interval $[0,2]$, but its inverse is not continuous and indeed $X$ is not homeomorphic to the interval $[0,2]$, because $[0,2]$ is connected and $X$ isn't.